我的问题是,我们会获得一系列活动及其开始和结束时间,然后我们必须找出我们可以做的最大活动。 这个问题非常普遍,我已经应用了贪婪的方法,但我的代码中出现了运行时分段错误,因此我将其发布在下面。
#include<stdio.h>
typedef struct event
{
long long int start_time;
long long int end_time;
long long int event_number;
}event;
long long int compare(const void *x, const void *y)
{
event *e1 = (event *)x, *e2 = (event *)y;
return (*e1).end_time - (*e2).end_time;
}
/* Given the list of events, our goal is to maximise the number of events we can
attend. */
int main()
{
int tests;
scanf("%d",&tests);
while(tests--)
{
long long int number_of_events;
//printf("enter no of activities");
scanf("%lld",&number_of_events);
event T[number_of_events];
long long int iter;
for(iter=0;iter<number_of_events;iter++)
{
// printf("enter start time and end time of %dth activity",iter);
scanf("%lld%lld",&T[iter].start_time,&T[iter].end_time);
T[iter].event_number = iter;
}
/* Sort the events according to their respective finish time. */
qsort(T,number_of_events,sizeof(event),compare);
long long int events[number_of_events]; // This is used to store the event
// numbers that can be attended.
long long int possible_events = 0; // To store the number of possible events
//Taking the first task
events[possible_events++] = T[0].event_number;
long long int previous_event = 0;
/* Select the task if it is compatable with the previously selected task*/
for(iter=1;iter<number_of_events;iter++)
{
if(T[iter].start_time >= T[previous_event].end_time)
{
events[possible_events++] = T[iter].event_number;
previous_event = iter;
}
}
//printf("Maximum possible events that can be attended are %d. They are\n",
// possible_events);
printf("%lld\n",possible_events);
/* list of selected activities
for(iter=0;iter<possible_events;iter++)
{
printf("%d\n",events[iter]);
}*/
}
return 0;
}
请帮助!! Thanx提前!!
答案 0 :(得分:0)
在最后的迭代中
for(iter=0;iter<possible_events;iter++)
您可能会使events
原因是,在possible_events++
的先前使用中,您通过possible_events
访问元素时,该变量的值将在访问之后变得比以前更大(因为后修复{ {1}}),所以如果你在每次迭代中调用评估,那么在循环结束后++
的值将为possible_events
你需要迭代到少一个:
number_of_events+1